A 100µF capacitor is connected in turn to two...?

Question

A 100µF capacitor is connected in turn to two different sinusoidal voltages of the same amplitude V=1.0V but different frequencies with f1 = 50Hz and f2 = 5kHz.

(i) Calculate the reactances X1 and X2.
(ii) By what factor is X1 greater then X2?
(iii) Calculate the amplitudes of the associated currents, I1 and I2.

Answer 1

capacitive impedance is:

Z = 1 / j(2 * pi * f * C)
Since there is no real component to the impedance, it is a pure reactance, X.

(i)
Simply plug in f1 and f2 into the equation to get the magnitude of the reactance. The phase of the impedance is 90 degrees.

(ii)
since this is a linear relationship (no 'squares' , 'squareroots', 'exponentials') so will the factor of X1 and X2 be a linear relationship. 50 Hz is 1/100th 5 kHz, so X1 will be 1/100th X2.

(iii)
i(t) = v(t) / X Ohms law... simple...

v(t) is 1.0 * sin(2* pi * f * t)
so i(t) will be 1/X * sin (2*pi*f*t)

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